Balance the following oxidation-reduction reactions, which occur in acidic solution, using the half-reaction method. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)

Br ‾(aq) + MnO4‾(aq) → Br2(l) + Mn2+(aq)

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To balance the oxidation-reduction reaction equation, we need to follow a few steps. Here's the step-by-step solution using the half-reaction method:

1. Divide the reaction into two half-reactions: oxidation and reduction.
Oxidation half-reaction: Br‾(aq) → Br2(l)
Reduction half-reaction: MnO4‾(aq) → Mn2+(aq)

2. Balance the atoms in each half-reaction, excluding H and O.
Oxidation half-reaction: 2Br‾(aq) → Br2(l)
Reduction half-reaction: MnO4‾(aq) → Mn2+(aq)

3. Balance the oxygen atoms by adding water molecules (H2O) to the side lacking oxygen.
Oxidation half-reaction: 2Br‾(aq) → Br2(l) + 2H2O(l)
Reduction half-reaction: MnO4‾(aq) + 4H2O(l) → Mn2+(aq)

4. Balance the hydrogen atoms by adding hydrogen ions (H+) to the appropriate side.
Oxidation half-reaction: 2Br‾(aq) + 2H2O(l) → Br2(l) + 4H+(aq) + 4e‾
Reduction half-reaction: MnO4‾(aq) + 4H2O(l) → Mn2+(aq) + 8H+(aq) + 5e‾

5. Balance the charges by adding electrons (e‾) to the appropriate side.
Oxidation half-reaction: 2Br‾(aq) + 2H2O(l) → Br2(l) + 4H+(aq) + 4e‾
Reduction half-reaction: MnO4‾(aq) + 4H2O(l) + 5e‾ → Mn2+(aq) + 8H+(aq) + 5e‾

6. Multiply each half-reaction by the necessary coefficients to make the number of electrons in each half-reaction equal.
Oxidation half-reaction: 2Br‾(aq) + 2H2O(l) → Br2(l) + 4H+(aq) + 4e‾
Reduction half-reaction: 5MnO4‾(aq) + 20H2O(l) + 5e‾ → 5Mn2+(aq) + 40H+(aq) + 5e‾

7. Combine the half-reactions and cancel out the electrons to obtain the balanced equation.
Overall balanced equation: 2Br‾(aq) + 5MnO4‾(aq) + 16H+(aq) → Br2(l) + 5Mn2+(aq) + 8H2O(l)

The balanced equation in acidic solution is:
2Br‾(aq) + 5MnO4‾(aq) + 16H+(aq) → Br2(l) + 5Mn2+(aq) + 8H2O(l)